Question
Question: The period of a function \[{2^{\left\\{ x \right\\}}} + \sin \pi x + {3^{\left\\{ {\dfrac{x}{2}} \ri...
The period of a function {2^{\left\\{ x \right\\}}} + \sin \pi x + {3^{\left\\{ {\dfrac{x}{2}} \right\\}}} + \cos 2\pi x (where \left\\{ x \right\\}denotes the fractional part of x) is
A.2
B.1
C.3
D.None of these
Solution
In this question first of all we calculate the period of all the individual terms of the given function. After that, to calculate the period of a function take Least common multiple (LCM) of period of all the terms. Which will be the period of our given function.
Complete step-by-step answer:
We have given f\left( x \right) = $$$${2^{\left\\{ x \right\\}}} + \sin \pi x + {3^{\left\\{ {\dfrac{x}{2}} \right\\}}} + \cos 2\pi x
As we know that,
Period of fractional part is always 1
⇒x=1
⇒Period of {2^{\left\\{ x \right\\}}}= 1 …(1)
For period of {3^{\left\\{ {\dfrac{x}{2}} \right\\}}}
Consider, 2x=1
\Rightarrow $$$$x = 2
⇒Period of {3^{\left\\{ {\dfrac{x}{2}} \right\\}}}=2 …(2)
Period of sine function is 2π
⇒ Period of sinπx=2π
⇒ Period of πx=2π
⇒x=π2π
⇒ Period of sinπx is x=2 …(3)
Period of cosine function is 2π
For period of cos2πx
Consider, 2πx=2π
⇒x=2π2π
⇒x=1 …(4)
For calculating the period of any function we have to take LCM of the period of all the terms given in the above function.
So, period of {2^{\left\\{ x \right\\}}} + \sin \pi x + {3^{\left\\{ {\dfrac{x}{2}} \right\\}}} + \cos 2\pi x is LCM (1,2,2,1) …(from 1,2,3 and 4)
Thus, period of {2^{\left\\{ x \right\\}}} + \sin \pi x + {3^{\left\\{ {\dfrac{x}{2}} \right\\}}} + \cos 2\pi x is 2.
Hence, option A. 2 is the correct option.
Note: Period of a function: The distance between the repetition of any function is called the period of the function.
For a trigonometric function the length of one complete cycle is called a period.
Fraction part function is a periodic function with period 1.